Плизз на помощь!!11!1!1

Решите задачу:

$a)\; \frac{a^2-b^2-c^2-2bc}{b^2-c^2-a^2+2ac}=\frac{2}{3}\; ,\; \; \frac{a^2-(b+c)^2}{b^2-(a-c)^2}=\frac{2}{3}\; ,\; \; \frac{(a-(b+c))(a+(b+c)))}{(b-(a-c))(b+(a-c))}=\frac{2}{3}\; ,\\\\ \frac{(a-b-c)(a+b+c)}{(b-a+c)(b+a-c)}=\frac{2}{3}\; \; \Rightarrow \; \; \; \frac{a+b+c}{b+a-c}=\frac{2(b-a+c)}{3(a-b-c)}\\\\b)\; \; \frac{c^2-a^2-b^2-2ab}{a^2-b^2-c^2+2bc}=\frac{8}{27}\; ,\; \; \frac{c^2-(a+b)^2}{a^2-(b-c)^2}=\frac{8}{27} \; ,\; \; \frac{(c-(a+b))(c+(a+b))}{(a-(b-c))(a+(b-c))}=\frac{8}{27} \; ,$

$\frac{(c-a-b)(c+a+b)}{(a-b+c)(a+b-c)}=\frac{8}{27}\; \; \Rightarrow \; \; \; \frac{a+b+c}{a+b-c}=\frac{8(a-b+c)}{27(c-a-b)}\\\\c)\; \; \frac{2(b-a+c)}{3(a-b-c)}=\frac{8(a-b+c)}{27(c-a-b)}\; \; \Rightarrow \; \; \; \frac{2\cdot 27\cdot (b+c-a)}{-3\cdot 8\cdot (b+c-a)}=\frac{a+c-b}{-(a+b-c)} \\\\ \frac{2\cdot 27}{3\cdot 8}=\frac{a+c-b}{a+b-c} \; \; \Rightarrow \; \; \; \sqrt{ \frac{a+c-b}{a+b-c} }=\sqrt{\frac{9}{4}}\\\\ \sqrt{ \frac{a+c-b}{a+b-c} } =\frac{3}{2}$

Плизз ** помощь!!11!1!1